(a 

I LIBRARY OF CONGRESS, t 

# # 






FES OF AMERICA.! 



A MANUAL 



W 



PROBLEMS ON THE GLOBES. 



DESIGNED AS AN ACCOMPANIMENT TO 

WILLIAMSON'S PATENT CONCENTRIC CELESTIAL 
AND TERRESTRIAL GLOBES. 



H. WILLIAMS OX, M.D., 

PRINCIPAL OF GRAMMAR SCHOOL NO. 53, NEW YORK ; INSTRUCTOR IN NATURAL 

PHILOSOPHY, ASTRONOMY, AND CHEMISTRY IN THE FEMALE NORMAL 

SCHOOL AND EVENING HIGH SCHOOL, NEW YORK. 







NEW YORK: 
HARPER & BROTHERS, PUBLISHERS, 

FRANKLIN SQUARE. 

186 8. 



Entered, according to Act of Congress, in the year 186S, by 

Harper & Brothers, 

In the Clerk's Office of the District Court of the United States for the 
Southern District of New York. 






PREFACE, 



The Author of the present little work, anticipating 
some difficulties in applying to his Concentric Celestial 
and Terrestrial Globes the rules for the solution of 
problems as laid down for ordinary globes in most 
works on Astronomy, felt the necessity of compiling a 
text-book of problems to accompany his globes by 
modifying and arranging to his purposes what has been 
so lucidly written by others. 

He therefore respectfully presents to the public this 
little " Manual of Problems on the Globes," and hopes 
that while he has made it suitable to his own especial- 
ly, he has rendered it applicable to other globes now 
in use. 

New York, March, 1868. 



r- 



CONTENTS, 



CHAPTEB I. PA6E 
Geographical and Astronomical Definitions, with Ex- 
planations 9 

CHAPTER II. 
Problems on the Terrestrial Globe 20 

CHAPTER III. 
Problems to be performed with the Terrestrial and 
Celestial Globes . 43 

CHAPTER IV. 
A Catalogue of the Constellations and Visible Stars, 
with their Locations, Representations, etc 54 

CHAPTER V. 
Problems to be performed by means of the Celestial 
Globe 75 



PROBLEMS ON THE GLOBES. 



CHAPTER I. 

GEOGEAPHICAL AND ASTRONOMICAL DEFINITIONS, WITH 
EXPLANATIONS. 

An artificial Globe is a miniature representation 
of the earth or heavens ; that which represents the 
earth is called the Terrestrial Globe, and that which 
represents the heavens the Celestial Globe. 

On the surface of the terrestrial globe, the various 
natural and political divisions of the earth are laid 
down ; and on the surface of the celestial globe, the 
constellations and fixed stars, all in their relative posi- 
tions. 

A Constellation is a group of fixed stars repre- 
sented under the name of some emblem or animal. 

In representing the relative positions of the earth 
and the heavens by means of " Williamson's Concen- 
tric Globes," the celestial globe is transparent, and sur- 
rounds the terrestrial; they turn on a common axle, 
and have a common centre, and are supported in a 
brass ring called the Celestial Brazen Meridian. 

The earth turns on her axis from west to east once 
in twenty-four hours, producing an apparent motion of 
the heavens from east to west ; this is called the Di- 
urnal motion. 

The diurnal motion of the earth may be represented 
A2 



10 PROBLEMS ON THE GLOBES. 

by causing the terrestrial globe to revolve from west 
to east by means of the milled head at the southern 
extremity of the axle, and the apparent daily motion 
of the heavens by causing the celestial globe to re- 
volve from east to west by means of the milled head 
at the northern extremity of the axle. 

The Axis of the earth is an imaginary line passing 
through her centre upon which she is supposed to re- 
volve, and is represented on the terrestrial globe by 
the axle on which it revolves. 

The Terrestrial Poles are two imaginary points 
where the axis is supposed to cut the surface of the 
earth : that which is north is called the North Pole, 
and that south, the South Pole. These points are 
seen on the terrestrial globe where the axle penetrates 
the surface. 

The Celestial Poles are two imaginary points in 
the heavens exactly over the terrestrial poles, around 
which the heavenly bodies appear to move, and are 
seen on the celestial globe w^here the axle penetrates 
its surface. 

Certain Great and Small Circles are drawn on 
the celestial and terrestrial globes representing circles 
supposed to have been drawn on the celestial and ter- 
restrial spheres for the purpose of measurement, and 
of determining the position of places and heavenly 
bodies. 

Great Circles are those whose planes divide the 
globe into two equal parts. 

Small Circles are those whose planes divide the 
globe into two unequal parts. 

The great circles drawn on the globes are the Equa- 
tor, the Ecliptic, the Meridian Circles, and the Co- 
lures. 



PROBLEMS ON THE GLOBES. 11 

The small circles drawn on the globes are the Trop- 
ics, the Polar Circles, and Parallels. 

The Equator is a great circle which divides the 
globe into Northern and Southern Hemispheres, 
and is every where equidistant from the poles. 

When applied to the heavens, or the celestial globe, 
the equator is called the Equinoctial. 

The Ecliptic is a great circle whose plane cuts that 
of the equator at an angle of 23° 28', and represents 
the plane of the earth's orbit. 

Meridian Circles are great circles which pass 
through the poles, crossing the equator at right an- 
gles. 

A Meridian is half of a meridian circle, and extends 
from pole to pole. 

The Tropics are small circles parallel to the equa- 
tor, and 23° 28' from it. That which is north of the 
equator is called the Tropic of Cancer, and that south 
the Tropic of Capricorn. 

The Polar Circles are small circles parallel to the 
equator, and 23° 28' from the poles. That which is 
north of the equator is called the Artic Circle, and 
that south the Antarctic Circle. 

Parallels of Latitude are small circles parallel to 
the equator. 

The Celestial Brazen Meridian is divided into 
360°, numbered from 0° to 90° from the equator to the 
poles. 

The Terrestrial Brazen Meridian is a semicircle 
of brass extending from the north to the south pole of 
the terrestrial globe, and is divided into 180°, number- 
ed from 0° to 90° from the equator to the poles. 

The Zodiac on the celestial globe is a space extend- 
ing about 8° on each side of the ecliptic, within which 



12 PROBLEMS ON THE GLOBES. 

the apparent motions of the sun, moon, and greater 
planets are confined. 

The Colures are two imaginary great circles which 
intersect each other at the celestial poles, one of which 
crosses the ecliptic at the first degrees of Aries and 
Libra, and is called the Equinoctial Colure because 
the equinoctial also crosses the ecliptic at these points ; 
the other crosses the ecliptic at the first degrees of Can- 
cer and Capricornus, and is called the Solstitial Co- 
lure. 

The Hour Circle is a small circle of brass attached 
to the north pole of the celestial globe, and capable of 
being moved on the axle. It is divided into twenty- 
four equal parts or hours, and when moved in connec- 
tion with the globe, the celestial brazen meridian serves 
as an index. 

Around the north pole of the terrestrial globe an 
hour circle is mapped, and when this globe is revolved 
the terrestrial brazen meridian serves as an index. 

The Sensible Horizon is the circle bounding the 
view of the observer by the apparent meeting of the 
earth and sky. 

The Rational Horizon is a great circle whose plane 
passes through the centre of the earth parallel to the 
plane of the sensible horizon, and which separates the 
visible from the invisible hemispheres of the heavens. 

The Universal Horizon is a flat ring of brass, di- 
vided into two separate semicircles, which are suspend- 
ed on two forked arms between the terrestrial and ce- 
lestial globes, and which represents the rational hori- 
zon. It is divided into three concentric circles, the ex- 
terior of which is marked Amplitude, and contains 
360°, numbered from the east and west points to the 
poles from 0° to 90°. 



PROBLEMS ON THE GLOBES. 13 

The central circle is marked Azimuth, and also con- 
tains 360°, and is numbered from the north and south 
points toward the east and west from 0° to 90°. 

The interior circle is divided into 32 parts, which 
represent the points of the compass ; these are subdi- 
vided into halves and quarters. 

The Cardinal Points of the horizon are East, 
West, North, and South. 

The Mariner's Compass attached to the globes, and 
which represents the sensible horizon, is a circular brass 
box containing a card divided into 32 points, and fixed 
on a magnetic needle which points toward the north. 

The Quadrant of Altitude is a thin, flexible strip 
of brass, divided upward from 0° to 90°, and downward 
from 0° to 18°, and capable of being screwed to the 
celestial brazen meridian. The upper divisions are 
used for determining the distances between heavenly 
bodies, their altitudes, etc., and the lower divisions for 
finding the beginning, duration, and end of twilight. 

Zones are five spaces or belts, into which the sur- 
face of the earth is divided by the two tropics and 
the two polar circles. The Torrid Zone is that space 
included between the two tropics, and is 46° 56' in 
width. 

That included between the tropic of Cancer and the 
arctic circle is called the North Temperate Zone, 
which is 43° 4' in width, and that between the tropic 
of Capricorn and the antarctic circle the South Tem- 
perate Zone, which also is 43° 4' in width. 

The space between the arctic circle and the north 
pole is called the North Frigid Zone, and that be- 
tween the antarctic circle and the south pole the 
South Frigid Zone, and are each 46° 56' in width. 

Latitude on the terrestrial globe is the distance 



14 



PROBLEMS ON THE GLOBES. 



north or south of the equator; on the celestial globe, 
the distance north or south of the ecliptic, 

The greatest latitude of a place is 90°. The sun, be- 
ing always on the ecliptic, can have no latitude. The 
greatest latitude of a planet is 8°, and of a star 90°. 

The latitude of a place is measured on the terrestrial 
brazen meridian ; of a heavenly body, on the quadrant 
of altitude toward the pole of the ecliptic. 

Longitude of a place is its distance east or west of 
the first meridian, measured on the equator half way 
around the globe ; of a heavenly body, its distance from 
the first point of Aries, measured on the ecliptic east- 
ward around the globe. 

The greatest longitude of a place is 180°; of a heav- 
enly body, 360°. 

The length of a degree of latitude is in every place 
60 geographical miles, but the length of a degree of 
longitude varies from 60 geographical miles at the 
equator to nothing at the poles. 



Table showing the Length of a Degree of Longitude on 
any Parallel of Latitude between the Equator and the 
Poles. 



Deg. 

of 


Geograph- 


Deg. 
of 


Geograph- 


Deg. 
of 


Geograph- 


Deg. 
of 


Geograph- 


Lat. 


ical Miles. 


Lat. 


ical Miles. 


Lat. 


ical Miles. 


Lat. 


ical Miles. 


I 


5 9-99 


1 1 


58. 9 


21 


56.01 


3l 


51.43 


2 


69.96 


12 


58. 69 


22 


55.63 


32 


5o.88 


3 


59.92 


13 


58.46 ■ 


23 


55.23 


33 


5o.32 


4 


59.85 


i4 


58.22 


24 


54.8i 


34 


49.74 


5 


59.77 


i5 


57.96 


•2 5 


54.38 


35 


49-i5 


6 


5 9 .6 7 


16 


57.67 


26 


53. 9 3 


36 


48.54 


7 


59.55 


17 


5 7 .38 


27 


53.46 


37 


47.92 


8 


59.42 


18 


57.06 


28 


52.98 


38 


47.28 


9 


59.26 


l 9 


56. 7 3 


29 


52.48 


3 9 


46.63 


IO 


59.09 


20 


56.38 


3o 


5i .96 


4o 


45.o6 



PROBLEMS OX THE GLOBES. 



15 



Pr- 
of 


Geograph- 


Deg. 

of 


Geograph- 


Deg. 

of 


Geograph- 


Deg. 

9f 
Lat. 


Geograph- 


Lai 


ical Miles. 


Lat. 


ical Miles. 


Lat. 


ical Miles. 


ical Miles. ■ 


4i 


45.28 


54 


35 .27 


67 


23.44 


79 


11.45 


42 


44. 5 9 


55 


34.4i 


68 


22.48 


80 


10.42 


43 


43.88 


56 


33.55 


69 


21 . 5o 


81 


9.39 


44 


43.16 


5 7 


32.68 


7° 


20. 52 


82 


8,35 


45 


42.34 


58 


3i.8o 


7 1 


19 = 53 


83 


7 .3i 


46 


4i.63 


5 9 


3o. 90 


7 2 


i8.54 


64 


6.27 


47 


4o . 92 


60 


3o ( 00 


73 


17.54 


85 


5.23 


48 


4o. 1 5 


61 


29 . 09 


74 


i6.54 


86 


4.19 


49 


3 9 . 36 


62 


28.17 


73 


i5 .53 


87 


3.i4 


5o 


38.57 


63 


27.24 


76 


i4.52 


88 


2 .00 


5i 


37.76 


64 


26.3o 


77 


i3. 5o 


89 


1 .o5 


52 


36.94 


65 


25.36 


78 


12.47 


90 


0.00 


53 


36.ii 


66 


24.40 











Declixatiox is the distance of a heavenly body 
north or south from the equinoctial. 

The greatest declination of the sun is 23° 28'; of a 
planet, about 31 J c ; and of a star, 90°. 

If the heavenly body be north of the equinoctial, it 
has xorth declixatiox; and if south, south decli- 

X ATI OX. 

The Ele.vatiox of the Pole at airy place is the 
height of the pole above the horizon, measured on the 
celestial brazen meridian. It is always the same num- 
ber of degrees as the latitude of the place. 

Right Ascexsiox - of a heavenly body is its distance 
east from the first point of Aries, measured on the 
equator. 

Declination and right ascension in the heavens cor- 
respond to latitude and longitude on the earth. 

The Zexith is that point in the heavens directly 
overhead of the observer, and is 90° from every point 
in the horizon. 

The Nadir is that point in the heavens directly op- 
posite to the zenith. 



16 PROBLEMS ON THE GLOBES. 

Vertical Circles are circles supposed to be drawn 
through the zenith and nadir of any place, cutting the 
horizon at right angles. 

The Prime Vertical is that which passes through 
the east and west points of the horizon, cutting the 
meridian at right angles. 

The Azimuth of a heavenly body is that arc of the 
horizon intercepted between a vertical circle passing 
through the object, and the north or south points of 
the horizon. 

Amplitude of a heavenly body is that arc of the 
horizon comprehended between an object at rising or 
setting, and the east or west points of the horizon. 

Parallels of Celestial Latitude are small cir- 
cles drawn on the celestial globe parallel to the eclip- 
tic. 

The Altitude of a heavenly body is an arc of a 
vertical circle intercepted between that body and the 
horizon. 

When the heavenly body is on the meridian, this arc 
is called the Meridian Altitude. 

The Zenith distance of a heavenly body is an arc 
of a vertical circle comprehended between that body 
and the zenith. 

The Polar distance of a heavenly body is an arc 
of a meridian comprehended between that body and 
the equinoctial pole. 

The Amphiscii are the inhabitants of the torrid zone 
so called, because their shadows at noon point north 
or south at different seasons of the year. 

The Heteroscii are the inhabitants of the temperate 
zone so called, because their shadows at noon always 
point in the same direction. 

The Periscii are those who inhabit the frigid zones, 



PROBLEMS OX THE GLOBES. 17 

because their shadows during a revolution of the earth 
on her axis point to all points of the horizon. 

The Antozci are those who have the same degree 
of longitude and equal degrees of latitude, but one 
north and the other south of the equator. 

The Peeiozci are those who have the same latitude, 
but opposite longitudes. 

The Axtipodes are those who live diametrically op- 
posite to each other. 

Ckepusculum, or Twilight, is that faint light which 
we perceive before the sun rises and after he sets. It 
is seen in the evening until the sun has descended 18° 
below the horizon, and in the morning when the sun is 
again within 18° of the horizon. 

The ecliptic and zodiac are divided into twelve equal 
parts called Signs, which bear the names of the con- 
stellations which formerly corresponded with them. 
Each sign is divided into thirty equal parts or degrees. 

The signs of the zodiac do not now correspond with 
the constellations of the zodiac, the difference between 
them being about thirty degrees. The sign Aries com- 
mences where the ecliptic is crossed by the equinoctial 
colure,. while the constellation Aries commences about 
thirty degrees more to the east. 

Owing to the motion of the earth in her orbit, the 
sun has an apparent motion in the heavens passing 
through the signs of the zodiac in succession from 
west to east at the rate of one degree nearly each day. 

The names of the signs, and the days on which the 

sun enters them, are as follows : 

Sjjring Signs. 1 

T Aries, the Ram, March 21st. i x ^ Al c . 

w rr. ^ -o n a -i , r,^ r -N orthern bigns. 

tf Taurus, the Bull, April 19th. f ° 

LT Gemini, the Twins, May 20th. J 



18 



PROBLEMS ON THE GLOBES. 



,1 



Northern Signs. 



Summer Signs. 
£s> Cancer, the Crab, June 21st. 
ft, Leo, the Lion, July 22d. 
TIE Virgo, the Virgin, August 22d. 

Autumn Signs. 
=g= Libra, the Balance, September 23d. 
HI Scorpio, the Scorpion, October 23d. 
£ Sagittarius, the Archer, November 22c*. i Southern 

. Winter Signs. Signs. 

Y3 Capricornus, the Goat, December 21st. 
ffl Aquarius, the Waterman, January 20th. 
^ Pisces, the Fishes, February 19th. 

The first degrees of the signs Cancer and Capricorn- 
us are called the Solstitial points, because the sun 
when at these points appears for some days to stand 
still — that is, he does not increase or diminish his dec- 
linations : the former is called the Summer Solstice, 
and the latter the Winter Solstice. 

The Angle of Position between two places on the 
earth is an angle at the zenith of one of the places 
formed by the meridian of that place and a vertical 
circle passing through the other place, measured on 
the horizon from the elevated pole toward the vertical 
circle. 

The angle of position of a heavenly body is an angle 
formed by two great circles intersecting each other in 
the place of the heavenly body, the one passing through 
the pole of the ecliptic, the other through the equinoc- 
tial pole. 

The Diurnal Arc is the arc described by a heaven- 
ly body from its rising to its setting. 

The Nocturnal Arc is the arc described by a heav- 
enly body from its setting to its rising. 



PROBLEMS ON THE GLOBES. 19 

Aberration is an apparent motion of celestial bod- 
ies, caused by the earth's motion in her orbit, combined 
with the progressive motion of light. 

POSITIONS OF THE SPHERE. 

A Right Sphere is that position of the earth where 
the equinoctial passes through the zenith and nadir, 
the poles being at the horizon. It is so called because 
the parallels of latitude cut the horizon at right angles. 
The inhabitants who live at the equator enjoy this po- 
sition of the sphere. In this position it will be ob- 
served that the diurnal and nocturnal arches of any 
heavenly body are equal ; consequently the days and 
nights are of equal lengths. 

The Parallel Sphere is that position of the earth 
where the parallels of latitude are parallel to the hori- 
zon. The inhabitants (if any) "who live at the poles 
have this position. The sun appears above the hori- 
zon for six months together. 

The Oblique Sphere is that position of the earth 
where the parallels of latitude cut the horizon oblique- 
ly. All the inhabitants of the earth, except those at 
the equator and at the poles (if any), have this posi- 
tion. It will be observed that in this position the 
parallels are each divided by the horizon into two un- 
equal parts ; hence the days and nights are of unequal 
lengths, except when the sun is on the equinoctial. 

When either pole is elevated above the horizon, there 
is a certain circle, having this pole for its centre, and 
its altitude for radius, wherein the stars do not sink be- 
neath the horizon ; this is called the Circle of Per- 
petual Apparition. 

A circle of precisely similar dimensions surrounds 
the depressed pole, wherein the stars do not rise above 



20 PROBLEMS ON THE GLOBES. 

the horizon, which is called the Circle of Perpetual 
Occultation. 

The Analemma is a table usually in the shape of 
the figure 8 on the terrestrial globe, extending from 
the tropic of Cancer to the tropic of Capricorn, used 
for ascertaining the sun's declination on any day in 
the year. 

The Circle of Illumination is that great circle 
which divides the hemisphere presented to the sun 
from the hemisphere which is deprived of his light. 



CHAPTER II 

PROBLEMS ON THE TERRESTRIAL GLOBE. 

PROBLEM I. 
To find the latitude of any given place. 

Rule. — Cause the terrestrial globe to revolve by 
turning the milled screw at the south pole of the celes- 
tial globe until the given place is brought under the 
graduated side of the terrestrial brazen meridian ; the 
degree above the place is the latitude. 

When a place is north of the equator, it is in north 
latitude ; when south, south latitude. 

Examples, — Find the latitude of the following 



places : 




London, Arts. 51° 31' N. 


Washington, Ans.38° 53' N". 


New York, * 40° 43' K 


Madrid, " 40°25'K 


Calcutta, " 22° 33' N". 


Paris, " 48°50'N. 


Valparaiso, " 33° 2' S. 


Vienna, " 48°12'N. 



PROBLEMS ON THE GLOBES. 21 



PROBLEM II. 

To find those places ichich have the same latitude as any 
given place. 

Rule. — Find the latitude of the given place by 
Problem I. Cause the terrestrial globe to revolve, 
and all places passing under this latitude on the ter- 
restrial brazen meridian will be those required. 

Note. — All places in the same latitude have the 
same length of day and night, and the same seasons of 
the year. 

Find the principal places which have the same or 
nearly the same latitude as Canton. 

Ansioer. Calcutta, Havana. 

Name some places where the length of each day in 
the year corresponds with that at New York. 

Answer. Naples, Constantinople. 

Name some places where the seasons correspond 
with those at Edinburg. 

Ansioer. Copenhagen, Moscow. 

PROBLEM III. 
To find the longitude of any given place. 

Rule. — Bring the given place under the graduated 
side of the terrestrial brazen meridian ; it's longitude 
will be found on the equator where it is crossed by 
this meridian. 

When the given place is east of the meridian which 
passes through Greenwich, near London, it is in east 
longitude ; when west, in west longitude. 

Find the longitude of the following places : 



22 PROBLEMS ON THE GLOBES. 



London, 


Answer. 0° 0' E. orW. 


New York, 




74° 0' W. 


Calcutta, 




" 88° 19' E. 


Valparaiso, 




71° 42' W. 


Washington, 




" 77° 3' W. 


Madrid, 




" 3° 45' W. 


Paris, 




2° 20' E. 


Vienna, 




" 16° 23' E. 


at is the greatest 


longitude a place can h 




PROBLEM IV. 



To find those places which have the same longitude as a 
given place. 

Rule. — Bring the given place under the graduated 
side of the terrestrial brazen meridian, and all places 
under this meridian from north pole to south pole will 
be those required. 

Note. — It will be noon at the same time at all places 
situated under the same meridian. 

Name a place where the longitude is the same, or 
nearly the same, as that of Stockholm. Answer. Dant- 
zic, Cape of Good Hope. 

PROBLEM V. 
To find the latitude and longitude of any given place. 

Rule.- — Bring the given place to the graduated side 
of the terrestrial brazen meridian ; the latitude will be 
found on the meridian over the place, and the longitude 
on the equator where it is crossed by the meridian. 



PROBLEMS ON THE GLOBES. 23 



PROBLEM VI. 

To find a place on the globe, its latitude and longitude 
being given. 

Rule. — Bring that longitude on the equator which 
corresponds with the longitude of the given place to 
the graduated side of the terrestrial brazen meridian, 
and under the given latitude on the meridian will be 
found the place required. 

What places have the following latitudes and lon- 
gitudes ? 

Lat. Long. 

50° 4' N. 5° 45' W. Answer. Land's End. 

48° 12' K 16° 23' E. " Vienna. 

52° 4'N. 4°19'E. " The Hague. 

19° 26' 1ST. 103° 46' W. " Mexico. 

42° 21' K 11° 4' W. " Boston. 

59° 21' N. 18° 4' E. ' " Stockholm. 

Where on the earth is there neither latitude noi 
longitude ? Answer. Where the first meridian crosses 
the equator. 

PROBLEM VII. 

- To find the difference of latitude between any two places. 

Rule. — Find the latitude of each place by Problem 
I. If both be north or both south, subtract the less 
from the greater, and the remainder will be the differ- 
ence of latitude. If one be north and the other south, 
their sum will be the difference of latitude. 

• Find the difference of latitude between the follow- 
ing places: 

New York and New Orleans, 40° 43' N. — 29°> 58' K 
~\0°Ab\A?is. ' 



24 PROBLEMS ON THE GLOBES. 

London and the Cape of Good Hope, 51° 31' N. + 34° 
22' S. = 85° 53', Arts. 

Quebec and Rio Janeiro, 46° 49' N.+22° 55' S.=z69° 
44', Arts. 

San Francisco and Valparaiso, 37° 48' N. + 33° 2' S. 
r=70° 50', Ans. 

St. Petersburg and New York, 59° 57' N. — 40° 43' N. 
= 19° 14', Ans. 

St. Helena and Jamaica, 15° 57' S. + 17° 40' K-33° 
37', Am. 

What two places on the earth have the greatest dif- 
ference of latitude ? Ans. The north and south poles. 

PROBLEM VIII. 

To find the difference of longitude between any two 
places. 

Rule. — Find the longitude of each place by Prob- 
lem III. If both be east or both west, their difference 
will be the difference of longitude. If one be east and 
the other west, their sum will be the difference of lon- 
gitude. 

Find the difference of longitude between the follow- 
ing places : 

London and New York, 0° 0'+ 74° 0' W. = 74° 0', Ans. * 

Constantinople and Washington, 28° 59' E. + 77° 3' 
W.=rl06° 2\Ans. 

Rome and Madrid, 12° 27' E. + 3° 45' W.=16° 12', 
Ans. 

Philadelphia and San Francisco, 75° 10' W. — 122° 26' 
W. = 47°16', Ans. 

Calcutta and New Orleans, 88° 19' E. + 90 7' W.= 
118° 26\ Ans. 

Dublin and Boston, 6° 20' W.-7l° 4' W.=64° 44', 
Ans. 



PROBLEMS OX THE GLOBES. 25 



PROBLEM IX. 

To find the difference of time between any two given 
places. 

Rule. — Find the difference of longitude between 
the given places by Problem VHX 

Every degree of difference of longitude is equal to 
four minutes difference of time, and every fifteen de- 
grees to one hour. Or, 

Multiply the difference of longitude in degrees and 
minutes by 4, and the product will be the difference of 
time in minutes and seconds. 

Find the difference of time between the following 
places : 

. Diff. of Long. Diff. of Time. 
h. m. sec. 

London and Washington, 77° 3'=: 5 8 12 Ans. 

Constantinople and N. York, 102° 59'= 6 51 56 " 
Calcutta and San Francisco, 200° 45' = 13 23 " 
Rio Janeiro and Lima, 33° 57' = 2 15 48 " 

St. Petersburg and Paris, 27° 59' = 1 51 56 " 

Boston and San Francisco, 116° 6'= 7 44 24 " 

PROBLEM X. 

The time at any given place being given, to find the 
time at any other given place. 

Rule. — Find the difference of time between the two 
given places by Problem IX. ; and if the place whose 
time is sought lie to the east of the place whose time 
is given, add the difference of time to the given time ; 
but if it lie to the west, subtract it. 

1. When it is 10 o'clock A.M. at New York, what 
time is it at London ? 

B 



26 PROBLEMS ON THE GLOBES. 

Longitude at New York, 74° 0' W. 

" " London, 0° 0' 

Difference of longitude, 74° 0'. 

Difference of time, 4 hours 56 minutes 9 seconds. 
Time at New York, 10 hours min. 
Difference of time, 4 " 56 " 

14 hours 56 min. = — 
2 hours 56 minutes after 12 o'clock. 
Answer. 56 minutes past 2 P. M. 

2. When it is midnight at Boston, what hour is it 
at Botany Bay ? 

Longitude of Boston, 71° 4' W. 

" " Botany Bay, 151° 15' E. 

Difference of longitude, 222° 19'. 

Difference of time, 14 hours 49 minutes 16 seconds. 

Subtracting 12 hours brings the time to noon of 
January 1st, and then subtracting 2 hours 49 minutes 
16 seconds from 12 hours gives 9 hours 10 minutes 44 
seconds. Answer. 10 minutes 44 seconds past 9 o'clock 
A.M., January 1st. 

3. When it is noon at Washington, what is the hour 
at the Cape of Good Hope ? 

Longitude of Washington, 77° 3' W. 

" " Cape of Good Hope, 18° 29' E. 

Difference of longitude, 95° 32'. 

Difference of time, 6 hours 22 minutes 8 seconds. 

Subtracting the difference in time from 12 o'clock, 
we get 5 hours 37 minutes 52 seconds. Answer. 37 
minutes 52 seconds past 5 A.M. 



PROBLEMS OX THE GLOBES. 2*7 



PROBLEM XL 

To find the Antoeei, Perwed^ and Antipodes to the in- 
habitants of any given place. 

Rule. — Bring the globes to the position of a right 
sphere, and turn the terrestrial globe until the given 
place is at the eastern half of the brazen horizon. 

Find on the amplitude circle the number of degrees 
the given place is north or south of the east point on 
the horizon. 

1st. The Axtoeci will be situated opposite that point 
which is numbered as many degrees on the amplitude 
circle on the other side of the east point of the horizon. 

2d. The Peeiceci will be situated opposite that point 
of the western half of the horizon which numbers as 
many degrees on the amplitude circle in the same di- 
rection from the west point of the horizon. 

3d. The Antipodes will be situated opposite that 
point of the western half of the horizon which num- 
bers as many degrees on the amplitude circle in the 
opposite direction from the west point of the horizon. 

Find the Antoeci of the following places, viz. : 

1. Falkland Islands. Ans. Eastern part of Labrador. 

2. Bermuda Islands. 3. Cairo, in Egypt. 
4. Jeddo. 5. Bourbon Island. 
6. Xew York. 7. Bombay. 

Find the Perioeci of the following places : 

8. St. Thomas Island. Ans. Hong Kong Island. 

9. Calcutta. 10. Luzon. 

11. Bermuda Islands. 12. Cape of Good Hope. 

Required the Antipodes to the following places : 
13. Marquesas Islands. Ans. Central part of Abys- 
sinia. 



28 PROBLEMS ON THE GLOBES. 

14. Dublin. 17. Isthmus of Suez. 

15. Moscow. 18. San Francisco. 

16. Christiana. 19. New Zealand. 

PEOBLEM XII. 

The difference of time between two places being given, 
to find their difference of longitude. 

Rule.— Reduce the difference of time to minutes, 
and divide by 4 ; multiply the remainder, if any, by 60, 
and divide the product by 4 ; the quotients will be the 
difference of longitude in degrees and minutes. 

Reduce the following differences of time to differ- 
ences of longitude : 

1. 4 hours 16 minutes. Ans. 64 degrees. 



2. 


21 " 


11 


3. 


14 " 


15 


4. 


3 " 


50 


5. 


6 " 


11 


6. 


1 " 


45 



7. The difference of time between London and New 
York is 4 hours 56 minutes ; what is their difference of 
longitude ? 

8. When it is 12 o'clock noon at Liverpool, it is 3 
minutes past 4 A.M. at San Francisco ; what is their 
difference of longitude ? 

PROBLEM XIII. 

A particular place, and the hour at that place, being 
given, to find all places on the globe where it is any 
other given hour. 

Rule. — Find the difference of time between the giv- 
en place and the place to be found. Reduce this dif- 



PBOORLEMg OK THE GLOBES, 29 

ference of time to difference of longitude by Problem 
XII. If the hour of the required places be earlier than 
that of the given place, the required places will lie so 
many degrees west of the given place ; but if later, to 
the east. 

1. When it is 12 o'clock noon at Hong Kong, at what 
places is it half past 6 A.M. ? 

2. TThen it is 3 P.M. at Port Philip, where is it mid- 
night ? 

3. TThen it is half past 5 P.M. at Constantinople, 
where is it 3 P.M. ? 

•4. TThen it is 10 A.M. at Edinburg, where is it 10 
P.M. ? 

5. TThen it is 7 A.M. at Florence, where is it 10 A.M. ? 

6. TThen it is noon at the Straits of Gibraltar, where 
is it 9 A.M. ? 

PROBLEM XIV. 

To find the smvs declination, and his longitude, or place 
on the ecliptic, for any given dag. 

Rule. — Bring the given day on the analemma under 
the graduated side of the terrestrial brazen meridian; 
the degree on the meridian immediately over it will be 
the sums declination for the given day. 

Turn the globe around until a point in the ecliptic 
passes under that degree on the meridian which shows 
the sums declination ; that point will be the sun's lon- 
gitude or place for the given day. 

If the given day be between March 21st and June 
21st. the sun's place will be between Aries and Cancer; 
if between June 21st and September 23d, it will be be- 
tween Cancer and Libra ; if it be between September 
23d and December 21st, between Libra and Capricor- 



30 PROBLEMS ON THE GLOBES. 

nus; and if it be between December 21st and March 
21st, it will be between Capricornus and Aries. 

1. What is the sun's longitude and declination on 
the 21st of June? Ans. 1st degree of® (Cancer); 
declination 23 degrees 28 minutes. 

Find the sun's longitude and declination for the fol- 
lowing days : 

2. January 4th. 3. February 11th. 

4. March 8th. 5. April 15th. 

6. May 28th. 7. June 19th. 

8. July 16th. 9. August 7th. 

10. September 1st. 11. October 4th. 

12. November 8th. 13. December 25th. 

PROBLEM XV. 

To rectify the terrestrial globe for any given latitude. 

Rule. — Bring that degree which corresponds with 
the given latitude, and which is marked on the lower 
half of the opposite side of the celestial brazen meridi- 
an, to the index on the stand. If the latitude be north, 
the north pole will be elevated ; and if south, depressed 
below the horizon a number of degrees corresponding 
with the given latitude. 

1. What is the altitude of the north polar star at 
Montreal? Ans. 45° 30'. 

2. What is the altitude of the north pole at Lisbon? 
Ans. 38° 42'. 

3. What is the altitude of the south pole at the Cape 
of Good Hope? Ans. 34° 22'. 



PROBLEMS ON THE GLOBES. 31 



PROBLEM XVI. 

Tlxe month and day of the month being given, to find 
all those places where the sun is vertical ; those places 
where the sun does not set, and those places where the 
sim does not rise on the given day. 

Rule. — Find the sun's declination for the given 
month. and day by Problem XIV., and observe the de- 
gree which corresponds with it on the terrestrial bra- 
zen meridian. Elevate the north pole if the declina- 
tion be north, and the south pole if south, to an alti- 
tude corresponding in degrees with the declination. 

Turn the terrestrial globe on its axis from west to east. 

1st. The sun will be vertical on that day to all those 
places which passes under the declination on the ter- 
restrial brazen meridian. 

2d. The sun does not set on that day to those places 
lying near the elevated pole, and which do not sink 
below the horizon. 

3d. The sun does not rise on that day to those places 
near the depressed pole which do not ascend above the 
horkon. 

1. At what places is the sun vertical on May 1st? 
Ans. At all those places having 15° north latitude. 

2. Where in the north frigid zone does the sun not 
set on August 10th? 

Ans. At all places north of 70° north latitude. 

3. Where in the south frigid zone does the sun not 
rise on the 1st of June. 

Ans. South of 68° south latitude. 

4. At what places is the sun vertical on Dec. 25th? 

5. " " " " March 3d? 

6. Where in the north frigid zone does the sun not 
set on April 3d ? 



32 PROBLEMS ON THE GLOBES. 

7. Where in the south frigid zone does the sun not 
rise on April 12th? 

9. Where in the south frigid zone does the sun not 
rise on May 8th ? 

PROBLEM XVII. 

To find those two days of the year on which the sun 
will be vertical at a given place in the torrid zone. 

Rule. — Turn the terrestrial globe until the analem- 
ma passes under the latitude of the given place on the 
terrestrial brazen meridian. The two days will be 
found on the analemma under that degree on the merid- 
ian corresponding with the latitude of the given place. 

1. In latitude 10° north, the sun is vertical on April 
15th ; on what other day of the year will he be vertical 
in the same latitude? Ans. August 28th. 

2. The sun is vertical at 20° south latitude on No- 
vember 21st; on what day will he again be vertical at 
the same place? Ans. January 22d. 

3. On what two days of the year will the sun be 
vertical in north latitude 15° ? Ans. May 2d and Au- 
gust 11th. 

PROBLEM XVIII. 

At any given place not in the frigid zones, the month 
and day of the mo?ith being given, to find what other 
day in the year is of the same length. 

Rule. — Find the given month and day on the ana- 
lemma ; the required month and day will be found op- 
posite to it, on the other side of the analemma. 

1. What other day in the year at London will be 
the same length as March 3d. Ans. October 10th. 



PROBLEMS ON THE GLOBES. 33 

2. At Trieste, what day in the year will correspond 
in length with February 28th? Ans. October 13th. 

3. What day at Jerusalem will correspond in length 
with December 25 th* ? Ans. December 19th. 

PEOBLEM XIX. 

The month, day, and hour at any place being given, to 
find that place at which the sun is vertical. 

Rule. — Find the sun's place on the ecliptic for the 
given month and day by Problem XIV., and note that 
degree on the terrestrial brazen meridian which shows 
his declination. Reduce the difference of time between 
the given hour and noon to degrees and minutes by 
Problem XII. Bring the given place to the terrestrial 
brazen meridian, and if the given time be earlier than 
noon, turn the terrestrial globe west, and if later, east, 
a number of degrees corresponding with the difference 
of time reduced to degrees, etc. ; the given place will 
be found under that degree on the meridian which 
shows the sun's declination. 

1. When it is 40 minutes past 6 in the morning at 
London on August 10th, where is the sun vertical? 
Ans. Lat. 15° 30' north, long. 80° east, southeastern part 
of Hindostan. 

2. When it is 3 P.M. at Charleston, May 8th, where 
is the sun vertical? Ans. Lat. 16° north, long. 124° 
51' west, in the Pacific Ocean. 

3. When it is 8 A.M. at Bombay on November 12th, 
where is the sun vertical? Ans. Lat. 18° south, long. 
133° 30' east, northern part of Australia. 

4. When it is 5 A.M. at San Francisco, March 30th, 
where is the sun vertical ? Ans. Lat. 3° 30' north, 
long. 16° 20' west, Atlantic Ocean. 

B 2 



34 PROBLEMS ON THE GLOBES. 



PROBLEM XX. 



The month, day, and hour at anyplace being given, to 
find, 1st, all those places on the earth where the sun is 
rising ; 2d, those places where the sun is setting ; 3 d, 
those that have noon / 4th, that particular place ichere 
the su?i is vertical ; 5th, those places that have morn- 
ing tioilight ; 6th, those places that have evening twi- 
light ; and, *lth, those places that have midnight. 

Rule. — Find the sun's declination by Problem XIV., 
and note it on the terrestrial brazen meridian. If the 
declination be north, elevate the north pole, and if 
south, the south pole, a number of degrees correspond- 
ing with the sun's declination. Bring the given place 
to the terrestrial brazen meridian. Reduce the differ- 
ence of time between the given hour and noon to de- 
grees and minutes by Problem XII. If the given hour 
be earlier than noon, turn the terrestrial globe west, 
and if later, east, a number of degrees corresponding 
with the difference of time reduced to degrees and 
minutes. 

1st. All those places along the western edge of the 
horizon have the sun rising. 

2d. All those places along the eastern edge of the 
horizon have the sun setting. 

3d. All those places under the terrestrial brazen me- 
ridian and above the horizon have noon. 

4th. That particular place under the sun's declination 
on the terrestrial brazen meridian has the sun vertical. 

5th. All those places below the western edge of the 
horizon, and within 18° of it, have morning twilight. 

6th. All those places below the eastern edge of the 
horizon, and within 18° of it, have evening twilight. 



PROBLEMS OX THE GLOBES. 35 

7th. All those places under that portion of the me- 
ridian circle passing through the given place which is 
below the horizon have midnight. 

1. When it is 7 A.M. at Washington on February 
17th, where is the sun rising, setting, etc? 

Ans. 1st. Sun rising centre of Greenland and Labra- 
dor, states of Ohio, Kentucky, and Louisiana, Gulf of 
Mexico, Yucatan, and Central America. 

2d. Sun setting in eastern part of Russia in Europe 
and of Independent Tartary, and central part of Hin- 
dostan. 

3d. Xoon west of Africa, central part of Spain, 1ST.E. 
part of Ireland. 

4th. Sun vertical at latitude 12° south, longitude 2° 
west. 

5th. Morning twilight at Melville Island, central part 
of British America, and along the Rocky Mountain 
ridge. 

6th. Evening twilight at central part of Liberia and 
Chinese Empire, Burmah, and at the western part of 
Australia. 

7th. Midnight at Kamtschatka, New Zealand, and 
Fiji Islands. 

PROBLEM XXI. 

To find the time of the sun's rising and setting p , and the 
length of day and night on any given day at any 
given place not in the frigid zone. 

Rule. — Find the sun's declination for the given day 
by Problem XIV. If it be north, elevate the north 
pole, and if south, the south pole, a number of degrees 
above the horizon corresponding with the declination, 
and bring the given place to the graduated side of 



36 PROBLEMS ON THE GLOBES. 

the terrestrial brazen meridian. On a parallel passing 
through the given place, ascertain how many degrees 
lie between the meridian and the eastern part of the 
horizon, and reduce these degrees to hours by Problem 
IX. 1 st. These hours subtracted from 1 2 will give the 
time of the sun's rising. 2d. These hours will be the 
time after 12 when the sun sets. 3d. Double these 
hours will be the length of the day. 4th. The length 
of the day subtracted from 24 will give the length of 
the night. 

1. At what time does the sun rise and set at Moscow 
on July 23d, and what is the length of the day and 
night ? 

Ans. Sun rises at 4 A.M. and sets at 8 P.M. ; length 
of day, 1 6 hours ; length of night, 8 hours. 

2. Required the length of the longest day and short- 
est night at Berlin. 

Ans. Longest day, 16 hours 12 minutes; shortest 
night, 7 hours 48 minutes. 

3. Required the length of the shortest day and long- 
est night at Lima. 

Ans. Shortest day, 11 hours 20 minutes; longest 
night, 12 hours 40 minutes. 

PROBLEM XXII. 

The length of the day at any place not in the frigid 
zones being given, to find the surfs declination, and 
the month and day of the month. 

Rule. — With the north and south poles of the ter- 
restrial globe at the horizon, bring the given place to 
the graduated side of the terrestrial brazen meridian ; 
tnen turn the globe eastward as many degrees as are 
equal to half the length of the day reduced to degrees 
by Problem XII. 



PROBLEMS ON THE GLOBESo Si 

Keep the terrestrial globe from revolving on its axis 
by screwing the milled head at the south of the axle. 
Elevate or depress the north pole until the given place 
is brought to the eastern edge of the horizon. The 
altitude of the north or south pole will correspond with 
the sun's declination in degrees. If the north pole be 
elevated the declination will be north, and if the south 
pole be elevated it will be south. Note this declina- 
tion on the terrestrial brazen meridian, and bring the 
analemma under it, when the months and days of the 
months will be seen. 

1. What two days of the year are each 14 hours in 
length at Rome, and what is the sun's declination on 
those days? 

Ans. May 9th and August 14th; declination, 17° 30' 
north. 

2. On what two days does the sun at New Orleans 
rise at 5 A.M. and set at 1 P.M. ? 

Ans. March 21st and September 23d. 

3. On what two days at St. Petersburg does the sun 
rise at 4 A.M. and set at 8 P.M. ? 

Ans. May 12th and August 2d. 

PROBLEM XXIII. 

To find when constant day begins and ends at any giv- 
en place in the north frigid zone. 

Rule. — Find the latitude of the given place by Prob- 
lem I., and subtract it from 90°. Find on the terres- 
trial brazen meridian a north declination correspond- 
ing in degrees with this difference; those days and 
months on the analemma which will pass under this 
declination will be the days required. 



38 PROBLEMS ON THE GLOBES. 

1. What is the length of the day, and when does it 
begin and when end at South Cape, Spitzbergen ? 

Arts. Length of constant day, 120 days; begins April 
8th, ends August 20th. 

2. Find the beginning, end, and duration of constant 
day at Cape Bathurst. 

Arts. Length of constant day, 65 days ; begins May 
20th, ends July 24th. 

3. Find the beginning, end, and duration of constant 
day at Disco Island. 

Ans. Length of constant day, 61 days; begins May 
22d, ends July 22d. 

PROBLEM XXIV. 

To find when constant night begins and ends, and its 
duration, at any given place in the north frigid zone. 

Rule. — Find the latitude of the given place by Prob- 
lem I., and subtract it from 90°. Find on the terres- 
trial brazen meridian a south declination correspond- 
ing in degrees with this difference ; those days on the 
analemma which will pass under this declination will 
be the days required. 

1. What is the length of constant night, and when 
does it begin and when end at the eastern extremity 
of Nova Zembla ? 

Ans. It begins on October 31st, and ends on Febru- 
ary 2 2d; duration, 115 days. 

2. Find the beginning, end, and duration of constant 
night at North Somerset Island. 

Ans. Constant night begins November 9th, and ends 
February 2d; duration, 85 days. 

3. Find the beginning, end, and duration of constant 
night at Cape Taimar, in Asia. 



PKOBLEMS ON THE GLOBES. 39 

Ans. Constant night begins October 27th, and ends 
February 15th; duration, 111 days. 

PKOBLEM XXV. 

To find the days, and number of days, on tchich the sun 
rises and sets every 24 hours, at any given place icith- 
in the north frigid zone. 

Rule. — Find the beginning, end, and duration of 
constant day at the given place by Problem XXTTT, 
Find also the beginning, end, and duration of constant 
night by Problem XXIV. The sun will rise and set 
every 24 hours on those days which are between those 
showing the end of constant day and the beginning of 
constant night, also on those between the end of con- 
stant night and beginning of constant day. 

Add together the lengths of constant day and con- 
stant night, and subtract the sum from 365 ; the re- 
mainder will show the number of days in the year on 
which the sun will rise and set every 24 hours at the 
given place. 

1. On how many days of the year does the sun rise 
and set at Greenland, in 70° north latitude ? 

Ans. Constant day from May 21st till July 24th. 

Constant night from November 21st till January 
22d; from July 24th till November 21st, 120 days; 
from January 22d till May 21st, 119. Total, 239 days. 

2. How many days in the year does the sun rise and 
set at the North Cape, in the Island of Maggeroe, lati- 
tude 71° 30' north? Ans. 215 days. 

3. On how many days in the year does the sun rise 
and set at the north of Spitsbergen ? Ans. 107 days. 



40 PROBLEMS ON THE GLOBES. 



PEOBLEM XXYI. 

To find in what degree of north latitude the sun begins 
to shine constantly on any given day between the 2\st 
of March and the 2\st of June / also in what degree 
in south latitude on any given day between the 23d 
of September and the 21st of December. 

Rule. — Find the sun's declination for the given day 
by Problem XIV., and subtract the number of degrees 
expressing it from 90 ; the difference will be the re- 
quired latitude. 

1. In what degree south latitude does the sun begin 
to shine constantly on the 18th .of October ? Ans. 80° 
south latitude. 

2. In what degree south latitude does the sun be- 
gin to shine without setting on November 20th? Ans. 
70° south latitude. 

3. In what degree north latitude does the sun begin 
to shine without setting on May 1st? Ans. 76° north 
latitude. 

PROBLEM XXVII. 

Any number of days not exceeding 186 being given, to 
find that latitude at which the sun does not set for the 
given number of days. 

Rule. — Count half the number of days backward or 
forward on the analemma from the 21st of June or 
the 22d of December, and bring the last day thus 
found under the terrestial brazen meridian, on which 
will be found the sun's declination for that day ; sub- 
tract this declination from 90°, and the remainder will 
be the latitude required. 



PROBLEMS ON THE GLOBES. 41 

1. In what degree does the sun shine for 150 days? 
Ans. 84° north or south latitude. 

2. In what latitude does the sun shine for 100 days? 
Ans. 76° north or south latitude. 

3. In what latitude does the sun shine for 60 days ? 
A?is. 10° 20' north or south latitude. 

PROBLEM XXVIII. , 

To place the terrestrial globe in a similar situation with 
respect to the sun as the earth is at the equinoctial, 
and at the summer and ivinter solstices, and thereby to 
show the comparative lengths of the days and nights 
at the different seasons of the year. 

Rule. — 1st. For the Equinoxes. — Since the sun at 
the equinoxes is vertical at the equator, rectify the 
globe for latitude 0° by Problem XV. The inner edge 
of the brazen horizon will represent the circle of illu- 
mination, which in this case extends from pole to pole, 
crossing the equator at right angles. And since all 
great circles on a globe mutually bisect each other, 
the circle of illumination will at this and all other sea- 
sons bisect the equator ; hence the diurnal arch will at 
the equator be at all times equal to the nocturnal arch, 
and the days and nights each 12 hours long. But at 
this season the circle of illumination bisects also every 
parallel from the north to the south pole; hence the 
days and nights are equal over the whole world. 

2d. For the Summer Solstice. — At the summer 
solstice the sun is vertical at 23° 28' north latitude. 
Rectify the globe for this latitude by Problem XV., 
when it will be observed, 1st. That some parallels near 
the south pole, viz., those south of 66 J° south latitude, 
do not come within the circle of illumination at anv 



42 PKOBLEMS OX THE GLOBES. 

part of the day, as may be illustrated by causing the 
globe to revolve on its axis. In these latitudes there 
is constant night. 2d. That those parallels between 
the equator and 66J° south latitude are unequally di- 
vided by the circle of illumination, making the noctur- 
nal arch greater than the diurnal arch; hence the 
nights are longer than the days in the ratio of the 
length of the nocturnal arch to the length of the diur- 
nal arch. 3d. That those parallels between the equa- 
tor and 661° north latitude are unequally divided by 
the circle of illumination, the diurnal arch being great- 
er than the nocturnal : consequently the days are lon- 
ger than the nights in the ratio of the diurnal to the 
nocturnal arch. 4th. That those parallels north of 
66^° north latitude are altogether within the circle of 
illumination, as may be illustrated by causing the 
globe to revolve on its axis. In these latitudes there 
is constant day. 

3d. For the Wixter Solstice. — The same explana- 
tions as those for the summer solstice will answer in 
this case, by simply substituting north for south, south 
for north, day for night, and night for day. 



CHAPTER m. 

PROBLEMS TO BE PERFORMED WITH THE TERRESTRIAL 
AXD CELESTIAL GLOBE. 

PROBLEM XXIX. 

To find the beginning, duration, and end of twilight at 
any given place on any given day. 

Rule. — Find the latitude of the given place by 
Problem I., and rectify the globes for that latitude by 
Problem XV. Find the sun's place for the given day 
on the ecliptic, and bring it under the celestial brazen 
meridian. Set the hour circle to 12, and screw the 
quadrant of altitude to the celestial brazen meridian 
at that degree corresponding with the latitude of the 
given place. Turn the celestial globe westward on its 
axis till the sun's place conies to the western edge of 
the horizon. The hours passed over by the index will 
show the time of the sun's setting, or the beginning of 
evening twilight. Continue the motion of the globe 
westward till the sun's place coincides with 18° on 
the quadrant of altitude below the horizon. The time 
passed over by the index from the time of the sun's 
setting will be the duration of evening twilight. 

1. Required the beginning, duration, and end of even- 
ing twilight at New Orleans on June 21st. Ans. Twi- 
light commences at 15 minutes past 7, and ends at 45 
minutes past 8 ; duration, 1 hour 30 minutes. 

2. Required the beginning, duration, and end of even- 
ing twilight at Washington on June 21st. Ans. Twi- 



44 PROBLEMS ON THE GLOBES. 

light commences 20 minutes past 7, and ends 15 min- 
utes past 9 ; duration, 1 hour 55 minutes. 

3. Required the beginning, duration, and end of even- 
ins; twilight at New York on June 21st. Arts. Twi- 
light commences at 7 hours 45 minutes, and ends 9 
hours 45 minutes ; duration, 2 hours. 

4. Required the beginning, duration, and end of even- 
ing twilight at Quebec on June 21st. Ans. Twilight 
commences at 8 hours, and ends 10 hours 45 minutes; 
duration, 2 hours 45 minutes. 

PKOBLEM XXX. 

To find the beginning, duration, and end of constant 
twilight and day at any given place. 

Rule. — To the latitude of the given place add 18°, 
and subtract the sum from 90. Find on the celestial 
brazen meridian a declination corresponding in degrees 
with this difference, and observe what two points on 
the ecliptic will pass under it. At that one of the two 
points which shows increasing declination will be 
found the day on which constant twilight begins, and 
at that one which shows decreasing declination will be 
found the day on which constant twilight ends. 

1. When do the inhabitants of London have constant 
day or twilight? Ans. From May 2 2d till July 21st, 
or 60 days. 

. 2. When do the inhabitants at the north of Scotland 
have constant twilight or day? Ans. From April 
28th till August 15th, or 109 days. 

3. When do those who live at Frobisher's Straits 
have constant day or twilight? Ans. From April 14th 
till August 29th, or 137 days. 



PKOBLEMS ON THE GLOBES. 45 

PROBLEM XXXI. 
To find the duration of twilight at the north pole. 

Rule. — Elevate the north pole 90°, and, 1st, observe 
what point in the ecliptic nearest to Libra passes un- 
der 18° on the celestial brazen meridian below the ho- 
rizon. Find on the ecliptic the month and day cor- 
responding with it. The time between September 21st 
and this day will be the duration of evening twilight. 
2d. Observe what point in the ecliptic nearest to Aries 
passes under 18° on the celestial brazen meridian be- 
low the horizon. Find on the ecliptic the month and 
day corresponding with it. The time between that 
day and the 21st of March will be the duration of 
morning twilight. 

1. What is the duration of twilight at the north 
pole, and what is the duration of dark night there ? 
Ans. Evening twilight commences at the 23d of Sep- 
tember and continues to the 13th of November, being 
51 days. Morning twilight commences January 29th 
and continues to March 21st, being 51 days. From 
September 23d till March 21st = 179 days, from which 
deduct 51 x 2 = 179 — 102 = 77 days total darkness. 

PROBLEM XXXII. 

To find that part of the equation of time which depends 
on the obliquity of the ecliptic. 

Rule. — Find the sun's place in the ecliptic by Prob- 
lem XIV., and bring it to the celestial brazen meridian. 
From the first degree of Aries to the graduated side 
of the celestial brazen meridian, count the number of 
degrees both on the ecliptic and the equinoctial. Sub- 



46 PEOBLEMS ON THE GLOBES. 

tract the less from the greater, and reduce to the differ- 
ence of time by Problem IX., which will give the equa- 
tion of time. If the number of degrees on the ecliptic 
exceed those on the equinoctial, the sun is faster than 
the clock; but if those on the equinoctial exceed 
those on the ecliptic, the sun is slower than the clock. 

Note. — The difference between a well-regulated 
clock and a sun-dial depends upon two causes, viz., the 
obliquity of the ecliptic, and the unequal velocity of 
the earth in her orbit. 

Rule. — Or, find the sun's place on the ecliptic by 
Problem XIV., and under the sign, and opposite the 
degree on the following table will be found the equa- 
tion of time required. 

1. What is the equation of time on the 4th of July? 
Arts. In Cancer, 15°; sun slower than clock 5 minutes 
9 seconds. 

2. What is the equation of time on November 1st? 
Arts. 9° in Scorpio; sun faster than clock 9 minutes 
36 seconds. 

3. What is the equation of time on January 12th? 
Arts. 22° in Capricornus ; sun slower than clock 1 min- 
utes 6 seconds. 



PROBLEMS ON THE GLOBES. 



47 



Table of Equation of Time. 



Sun faster than Clock in 






Aries. 


Taurus. 


Gemini. 


1st 


B5 

<u 
So 




°P 


a 


rr 


Quar. 




Libra. 


Scorpio. 


Sagittarius. 


3d 


Q 




-A- 


m 


# 


Quar. 


o 


min. sec. 


8 min. 24 sec. 


8 min. 46 sec. 


3o 


I 


' 


6 20 " 


8 44 35 44 


8 " 36 " 


29 


2 


4 


< 4o " 


8 44 45 44 


8 44 25 44 


28 


3 


1 4 


4 " 


8 4C 54 " 


8 44 i4 " 


27 


4 


1 4 


4 19 " 


9 44 3 44 


g a j u 


26 


5 


1 4 


- 3 9 « 


9 44 11 44 


7 " 49 " 


25 


6 


1 4 


4 5 9 « 


9 44 18 44 


7 44 35 44 


24 


7 


2 ' 


4 18 " 


9 44 24 " 


7 44 21 44 


23 


8 


2 4 


■ 3 7 « 


9 44 3i 44 


7 44 6 44 


22 


9 


2 4 


4 56 " 


9 44 36 44 


6 44 5i 44 


21 


10 


3 4 


4 16 " 


9 44 4i " 


6 44 35 44 


20 


ii 


3 < 


< 34 " 


9 44 45 44 


6 44 19 44 


r 9 


12 


3 < 


4 53 " 


9 44 49 " 


6 44 2 44 


18 


i3 


4 < 


t n « 


9 44 5i 44 


5 44 45 44 


17 


i4 


4 « 


c 29 u 


9 44 53 44 


5 a 27 44 


16 


i5 


4 ' 


4 4 7 " 


9 44 54 " 


5 44 9 44 


i5 


16 


5 4 


« 4 " 


9 44 55 44 


4 " 5o 44 


i4 


J 7 


5 ' 


' 21 " 


9 44 55 " 


4 u 3i 44 


i3 


18 


5 < 


6 38 " 


9 44 54 " 


4 " 12 a 


12 


x 9 


5 4 


' 54 " 


9 44 52 44 


3 44 5 2 44 


11 


20 


6 « 


1 10 " 


9 44 5o 44 


3 44 32 44 


10 


21 


6 < 


4 26 ll 


9 44 4 7 " 


3 44 12 44 


9 


22 


6 4 


4 4i " 


9 44 - 43 44 


2 44 5i 44 


8 


23 


6 < 


' 35 " 


9 44 38 44 


2 44 3o 44 


7 


24 


7 ' 


4 9 " 


9 44 33 44 


2 44 9 44 


6 


25 


7 ' 


4 23 " 


9 44 27 44 


1 44 48 " 


5 


26 


7 ' 


4 36 " 


9 44 20 44 


1 44 27 44 


4 


27 


7 ' 


« 4 9 " 


9 44 i3 44 


1 44 5 44 


3 


28 


8 < 


4 1 44 


9 44 5 44 


44 43 44 


2 


29 


8 < 


4 i3 44 


8 44 56 44 


44 22 44 


1 


3o 


8 ■ 


4 24 " 


8 44 46 " 


44 44 





2d 




Virgo. 


Leo. 


Cancer. 




Quar. 




TIB 


a 


£p 


9 

fcD 
O 


4th 




Pisces. 


Aquarius. 


Capricornus. 


Quar. 




x 


AW 


V3 


Q 


Sun slower than Clock in 



48 PROBLEMS ON THE GLOBES. 



PROBLEM XXXIII. 

To find the sun's meridian altitude for any given day 
at any given place. 

Rule. — Find the sun's declination for the given day 
by Problem XIV., and rectify the globe for a corre- 
sponding latitude by Problem XV. Find the latitude 
of the given place by Problem I. The number of de- 
grees on the celestial brazen meridian measured from 
a degree corresponding with the latitude to the near- 
est point of the horizon will be the required meridian 
altitude of the sun. 

1 . What is the sun's altitude at Vienna on Decem- 
ber 22d? Ans. 19°. 

2. How high is the sun from the horizon at Genoa 
on February 28th at 12 M. ? Ans. 27^°. 

3. What is the sun's meridian altitude on June 22d 
at Bermuda Island ? Ans. 81°. 

PROBLEM XXXIV. 

To find the sun's amplitude at any given place for any 
given day. 

Rule. — Find the latitude of the given place by 
Problem I., and rectify the globe for that latitude by 
Problem XV. Find the sun's place in the ecliptic for 
the given day by Problem XIV, and bring it on the 
celestial globe to the eastern edge of the horizon. The 
number of degrees on the horizon between the sun's 
place on the ecliptic and the east point on the horizon 
will be the required amplitude. 

1. What is the sun's amplitude at Land's End on 
August 8th? Ans. 23°. 



PROBLEMS ON THE GLOBES. 49 

2. At what point of the compass will the sun rise on 
September 23d at St. Helena? A?is. East; ampli- 
tude, 0°. 

3. At what point of the compass will the sun set on 
April 22d at Juan Fernandez Island ? Arts. East by 
north, 5° north; amplitude 1,* 5°. 

To find the sun's azimuth and his amplitude at any 
given place for any given day and hour. 

Rule. — Find the latitude of the given place by 
Problem I., and rectify the globe to that latitude by 
Problem XV. Screw the quadrant of altitude to the 
celestial brazen meridian over the latitude of the given 
place. Find the sun's place in the ecliptic for the 
given day by Problem XIV., and bring it, on the celes- 
tial globe, to the celestial brazen meridian ; then set 
the hour circle to 12 o'clock. If the given hour be be- 
fore noon, turn the globe eastward ; but if it be after 
noon, turn it westward as many hours on the hour cir- 
cle as are equal to the difference between the given 
hour and noon. Bring the graduated edge of the 
quadrant of altitude over the sun's place. Then, 1st. 
The number of degrees from the north or south point 
of the horizon to the graduated edge of the quadrant 
of altitude will be the required azimuth. 2d. The 
number of degrees on the quadrant of altitude from 
the horizon to the sun's place on the ecliptic will be 
the required altitude for the given day and hour. 

1. What is the sun's altitude and azimuth at Pekin 
on the 12th of August, at half past 10 A.M. ? Ans. 
Azimuth, 72° K ; altitude, 41°. 

2. What is the sun's altitude and azimuth on the 
28th of June, at half past 4 P.M., at Barbadoes ? Ans. 
Azimuth, 68° K ; altitude, 57°. 

C 



50 PROBLEMS ON THE GLOBES. 

3. What is the sun's altitude and azimuth at the 
Mauritius on May 1st at 2 P.M. ? Ans. Azimuth, 60° 
1ST.; altitude, 57°. 

PKOBLEM XXXV. 

The latitude of a place, the day of the tnonth, and the 
surfs altitude being given, to find the surfs azimuth, 
and the hour of the day. 

Rule. — Rectify the globe for the latitude of the 
given place by Problem XV., and screw the quadrant 
of altitude on the celestial brazen meridian over the 
latitude of the given place. On the celestial globe, 
find the sun's place in the ecliptic for the given day by 
Problem XIV., and bring it to the graduated side of 
the celestial brazen meridian. Set the hour circle to 
twelve, and turn the globe until the sun's place coin- 
cides with the given altitude on the quadrant. The 
hours passed over by the hour circle give the required 
time from noon, and the azimuth will be found on the 
horizon as in Problem XXXIV. 

1. At what hour on the 9th of March is the sun's al- 
titude 25° at London, and what is his? azimuth, obser- 
vation being made in the forenoon ? Ans. 8 hours 48 
minutes A.M. ; azimuth, 52-J° from the south. 

2. At what hour of the day at Quebec on the 20th 
of September is the sun's altitude 21°, and what is his 
azimuth, observation being made in the morning? 
Ans. 8 hours 30 minutes A.M. ; azimuth, 66° south. 

3. At what hour at Lisbon is the sun's altitude 30° 
on the 1 8th of May, and what is his azimuth, observa- 
tion being made in the afternoon ? Ans. 4 hours 30 
minutes P.M. ; azimuth, 88° from the north. 



PKOBLEMS ON TIIE GLOBES. 51 



PROBLEM XXXVI. 

TJie latitude of a place and the day of the month being 
given, to find at what hour the sun will be due east or 
icest. 

Eule. — Rectify the globe for the given latitude by 
Problem XV. Screw the quadrant of altitude on the 
celestial brazen meridian over the given latitude, and 
set the hour circle to twelve. On the celestial globe, 
find the sun's place in the ecliptic for the given day by 
Problem XIV., and bring it to the graduated side of 
the meridian. Bring the lower end of the quadrant of 
altitude to the east point of the horizon, and hold it in 
this position while the globe is turned on its axis un- 
til the sun's place comes to the graduated side of the 
quadrant of altitude. The number of hours passed 
over by the index, subtracted from twelve, will be the 
required hour when the sun will be due east, and this 
number added to twelve will be the required hour 
when the sun will be due west. 

1. At what hours will the sun be due east and west 
at Berlin on June 21st, and what will be his altitude? 
Ans. 7 hours 45 minutes A.M., and 4 hours and 15 
minutes P.M. ; altitude, 28°. 

2. At what hour will the sun be due east and west 
on May 30th at St. John's, Newfoundland, and what 
will then be his altitude from the horizon? A?is. 7 
hours 20 minutes A.M., and 4 hours 40 minutes P.M. ; 
altitude, 25°. 

3. At what hours on the 30th of August will the sun 
be due east and west at Port Philip, Australia, and 
what will then be his altitude ? Ans. 5 hours 45 min- 
utes A.M., and 6 hrs. 15 minutes P.M. ; altitude, —17°. 



52 PROBLEMS OX THE GLOBES. 



PROBLEM XXXVII. 

Given the sun's meridian altitude and the day of the 
month, to find the latitude of the place. 

Rule. — Find the sun's declination for the given day 
by Problem XIV., and if it be south, add to it the sun's 
altitude, but if north, subtract it. Subtract this sum 
or difference, as the case maybe, from 90°, and the re- 
mainder will be the required latitude of the place. 

1. What is the latitude of a place where the sun's 
altitude is 50° on March 30th ? Ans. 

2. What is the latitude of a place where the sun's 
altitude is 60° on May 20th ? Ans. 

3. What is the latitude of that place where the alti- 
tude of the sun is 70° on December 1st? Ans. { 

PROBLEM XXXYin. 

Ihe length of the longest day at any given place not 
loithin the polar circles being given, to find the lati- 
tude of that place. 

Rule. — If the place be north of the equator, bring 
the first point of the sign Cancer, or if south, of Capri- 
corn, to the graduated side of the celestial brazen me- 
ridian, and set the hour circle to twelve. Turn the ce- 
lestial globe westward until the index has passed over 
as many hours as equal half the length of the day. El- 
evate or depress the north pole until the first point of 
Cancer or Capricorn comes to the horizon. The alti- 
tude of the north or south pole will correspond with 
the latitude. 

1. In what degree north latitude is the length of the 
longest day sixteen hours ? 



PROBLEMS ON THE GLOBES. 53 

2. In what latitude south does the sun set on the 
22d of December at 7 o'clock? 

3. In what degree north latitude is the length of 
the longest day three times the length of the shortest 
night? 

PROBLEM XXXIX. 

Given the day of the month, and the su?i > s amplitude at 
sunrise, to find the latitude of the place of observa- 
tion. 

Rule. — Find the sun's place in the ecliptic by Prob- 
lem XIV., and bring it to the horizon. Elevate or de- 
press the north pole until the sun's place coincides 
with the given amplitude. The altitude of the poles 
will correspond in degrees with the required latitude. 

1. On the 2d of May the sun's amplitude toward the 
north was 18°. What was the latitude of the place of 
observation ? 

2. On the 21st of June the sun's amplitude was ob- 
served to be 24° toward the north. What was the 
latitude of the place of observation ? 

3. The sun's amplitude on December 2d was ob- 
served at sunrise to be 20° toward the south. What 
was the latitude of the place of observation ? 



54 PROBLEMS ON THE GLOBES. 



CHAPTER IV. 

A CATALOGUE OF THE CONSTELLATIONS AND VISIBLE 
STARS, TOGETHER WITH THEIR REfPRESENTATIONS, LO- 
CATIONS, ETC. 

The stars visible to the naked eye are divided into 
seven classes or magnitudes, those of the greatest bril- 
liancy being of the first magnitude, those next in bril- 
liancy of the second magnitude, those not quite so 
bright of the third, and so on ; but by the aid of the 
telescope many others are brought to view, and the 
gradation is carried on through these to the sixteenth 
magnitude. 

On the globe to which this catalogue is an accom- 
paniment those stars only of the first four magnitudes 
are laid down. 

Since 1603 astronomers denote the various stars in 
a constellation by the small Greek letters, the brightest 
being represented by a (Alpha), the next in brilliancy 
by /3 (Beta), and so on through the alphabet; and 
when these are exhausted they have recourse to the 
small Roman letters, and lastly to the numerals 1, 2, 3, 
etc. 

We here present the small letters of the Greek al- 
phabet for the benefit of those unacquainted with them. 

v Nu. t Tau. 

£ Xi. v Upsilon. 

o Omikron. <p Phi. 

7r Pi. x Chi. 

p Rho. ^ Psi. 

a Sigma. 0) Omega. 



a Alpha. 


t) Eta. 


/3 Beta. 


6 Theta. 


y Gamma. 


i Iota. 


S Delta. 


k Kappa. 


e Epsilon. 


X Lambda. 


K Zeta, 


^Mu. 



PROBLEMS ON THE GLOBES. 



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PROBLEMS ON THE GLOBES. 



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57 






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PROBLEMS ON THE GLOBES. Y5 



CHAPTER IV 

PROBLEMS PERFORMED BY MEAXS OF THE CELESTIAL 
GLOBE. 

PROBLEM XL. 

To find the right ascension and declination of a heav- 
enly body. 

Rule. — Bring the heavenly body under the gradu- 
ated side of the celestial brazen meridian, on which will 
be found the declination. The number of degrees on 
the equinoctial between the meridian and the first de- 
gree of the sign Aries will be the right ascension. 

1. What is the right ascension and declination ofAl- 
genib in Pegasus ? Ans. R. A., 1° ; declination, 14° N\ 

2. What is the right ascension and declination of 
Almak in Andromeda ? Ans. 28£° R. A. ; 41° declina- 
tion N". 

3. What is the right ascension and declination of 
Mira in Cetus ? Ans. R, A., 32^° ; declination, 4° S. 

4. What is the right ascension and declination of 
Antares in Scorpio? Ans. R. A., 245°; declination, 
26° S. 

PROBLEM XLI. 

To find the latitude and longitude of a heavenly body. 

Rule. — Screw the quadrant of altitude to the celes- 
tial brazen meridian over that ecliptical pole which is 
in the same hemisphere with the heavenly body. Move 



76 PROBLEMS ON THE GLOBES. 

the lower end of the quadrant until the heavenly body 
comes to its graduated edge, on which will be found 
the required latitude. The number of degrees on the 
ecliptic reckoned from the first degree of the sign Aries 
to the graduated edge of the quadrant of altitude will 
be the required longitude. 

1. What is the latitude and longitude of a Scheder 
in Cassiopeia ? Arts. Latitude, 61°; longitude, 44°. 

2. What is the latitude and longitude of y Bellatrix 
in Orion? Ans. Latitude, 5° N. ; longitude, 74°. 

3. What is the latitude and longitude of /3 Mirach 
in Andromeda? Ans. Latitude, 37°; longitude, 29°. 

4. What is the latitude and longitude of j3 Pollux in 
Gemini? Ans. Latitu.de, 31° ; longitude, 115°. 

PROBLEM XIII. 

The right ascension and declination of a heavenly body 
being given, to find its place on the globe. 

Rule. — Bring the given degree of right ascension to 
the graduated edge of the celestial brazen meridian, 
and under the given declination on it will be found the 
required heavenly body. 

1. What star has right ascension 11° 11', and declina- 
tion 59° 38' X. ? Ans. y in Cassiopeia. 

2. What star has right ascension 86° 13', and declina- 
tion 44 a 55' N". ? Ans. Menkalina in Auriga. 

3. What star has right ascension 113° 16', and decli- 
nation 28° 30' K ? Ans. Pollux in Gemini. 

4. What star has right ascension 48°, and declination 
49° N. ? Ans. Algenib in Perseus. 



PROBLEMS ON THE GLOBES. 77 



PROBLEM XLIII. 

The latitude and longitude of a heavenly body being 
given, to find its place on the globe. 

Rule. — Screw the quadrant of altitude to the celes- 
tial brazen meridian over that ecliptical pole which is 
in the same hemisphere with the required heavenly 
body, and bring that part of the graduated edge of the 
quadrant marked 0° to the required longitude on the 
ecliptic. The required heavenly body will be found 
under the given latitude on the quadrant. 

1. What heavenly body has latitude 12° X., and lon- 
gitude 8° ? Ans. Al^enib in Pegasus. 

2. What star has latitude 10° 4' X., and longitude 
107° 21' ? Ans. Castor in Gemini. 

3. What star has latitude 22° 52' X., and longitude 
. 78° 57' ? Ans. Capella in Auriga. 

4. What star has latitude 21° 6', and longitude 330° 
56'? Ans. Enif in Pegasus. 

PROBLEM XLIV. 

The day and hour, and the latitude of a place being 
given, to find what stars are rising, setting, adminat- 
ing, etc. 

Pule. — Rectify the globe for the given latitude by 
Problem XV. Find the sun's place in the ecliptic for 
the given day by Problem XIV., and bring it on the 
celestial globe to the graduated edge of the celestial 
brazen meridian. Set the hour circle to twelve ; if the 
given hour be before noon, turn the celestial globe 
eastward, but if past noon, westward on its axis until 
the index has passed over as many hours as the differ- 
ence between the given hour and noon. 



78 PROBLEMS ON THE GLOBES. 

.Then all heavenly bodies at the eastern half of the 
horizon will be rising, those at the western half set- 
ting, and those under the celestial brazen meridian 
will have culminated. All bodies above the horizon 
will be visible, and those under it invisible. 

If the globe be revolved on its axis, those stars 
which will not sink below the horizon are within the 
circle of perpetual apparition ; those which will not 
rise above the horizon are within the circle of perpet-' 
ual occupation. These circles surround the celestial 
poles, and have radii equal in degrees to the latitude 
of the given place. 

1. At 9 o'clock in the evening of August 12th at 
New York, what stars will be rising, what stars will 
be setting, and what stars will be culminating ? Ans. 
Rising, Capella in Auriga ; culminating, Salaphat in 
Lyra ; setting, Deneb in Leo Major. 

2. What stars do not set to the inhabitants of Van- 
couver's Island? Ans. Those north of 60° N.Dec., 
or those in the northern part of Hydra, Cepheus, Cas- 
siopeia, Camelopardalus, Ursa Major, and all in Ursa 
Minor. 

3. What stars do not rise to the inhabitants of the 
Madeira Islands ? Ans. All south of 60° S. Dec, such 
as those in Pava, Toucan, Hydrus, Piscis volans, Cha- 
meleon, etc. 

4. In what latitude must a person be in to have lost 
sight of Cassiopeia ? Ans. South of 60° south lati- 
tude. 



PROBLEMS ON THE GLOBES. 



PROBLEM XLV. 



The latitude of a place, the month, day, and hour being 
given, to place the globe in such position as will rep- 
resent the heavens at that time. 

Rule. — Rectify the globe for the given latitude by 
Problem XV., and set the globe due north and south 
by means of the mariner's compass. Find the sun's 
place in the ecliptic on the terrestrial globe by Prob- 
lem XIV., and find a declination corresponding with it 
on the celestial brazen meridian. Turn the celestial 
globe on its axis until the ecliptic passes under this 
declination, and set the hour circle to 12 o'clock. If 
the time be after noon, turn the globe westward, but if 
before noon, eastward until the index has passed over 
as many hours as the given time wants of noon. That 
half of the celestial globe which is above the horizon 
will represent the heavens at the given time. 

PROBLEM XL VI. 

To find when any given star will rise, culminate, or set 
at any given place, the day of the month being given. 

Rule. — Rectify the globe for the given latitude by 
Problem XV., and find the sun's place on the celestial 
globe as in the preceding problem, and bring it to the 
graduated side of the celestial brazen meridian. Set 
the hour circle to 12 o'clock. If the given star be be- 
low the horizon, turn the globe westward on its axis 
until it is brought to it : then will the index on the 
hour circle show the time from noon when it rises. 
Continue the motion westward until it is brought un- 
der the brazen meridian, when the time of its culmi- 



80 PROBLEMS ON THE GLOBES. 

nating will be found by means of the hour circle. Con- 
tinue the motion still farther westward until the star 
is again at the horizon. The time of its setting w^ill 
be found as before by means of the hour circle. 

1. At what time at Philadelphia on March 2d will 
Pollux rise, culminate, and set ? Ans. It rises at 1 
P.M., culminates at 8 hours 45 minutes P.M., and sets 
at 4 hours 30 minutes A.M., March 3d. 

2. At Montreal, when will Mira in Cetus rise, cul- 
minate, and set on February 3d? Ans. It rises at 11 
hours 30 minutes A.M., culminates at 5 hours 15 min- 
utes P.M., and sets at 8 hours 50 minutes P.M. 

3. At what time will 2 in Aquarius rise, culminate, 
and set at Gibraltar on August 12th? Ans. It rises 
at 9 P.M., culminates at 1 hour 30 minutes A.M., and 
sets at 1 A.M. 

PROBLEM XLYII. 

To find the rising and setting amplitude of a star at 
any given latitude. 

Rule. — Rectify the globe for the given latitude by 
Problem XV., and bring the given star to the eastern 
part of the horizon. The number of degrees between 
the star and the east point of the horizon will be its 
rising amplitude. Bring the star to the western part 
of the horizon. The number of degrees between the 
star and the west point of the horizon will be its set- 
ting amplitude. 

1. What is the rising and setting amplitude of Al- 
debaran at Lisbon? Ans. Rising amplitude, 21° north 
of east; setting amplitude, 21° north of west. 

2. What is the rising and setting amplitude of Arc- 
turus at Naples ? Ans. Rising amplitude, 26° north 
of east ; setting amplitude, 26° north of west. 



PROBLEMS ON THE GLOBES. 81 

3. What is the rising and setting amplitude of 
Regulus at Edinburg ? Ans. Rising amplitude, 22° 
north of east; setting amplitude, 22° north of west. 

PROBLEM XL VIII. 

To find the angle that any two stars icill subtend as 
seen by a spectator, or to find their distance from 
each other in degrees. 

Rule. — Lay the quadrant of altitude from one star 
to the other so that the degree marked may be over 
one of them. The number of degrees between them 
on the quadrant will be the angle they will subtend, 
or their distance in degrees apart. 

1. What is the distance in degrees between Nath in 
Taurus and Algieba in Leo ? Ans. 68°. 

2. What is the distance in degrees between Alwaid 
in Draco and Schedir in Cassiopeia? Ans. 57°. 

3. What is the distance in degrees between Alphe- 
ratz in Andromeda and Sheratan in Aries ? Ans. 26°. 

PROBLEM XLIX. 

Given the latitude of a place, to find the meridian alti- 
tude of a star. 

Rule. — Rectify the globe for the given latitude by 
Prob. XV. Bring the star to the celestial brazen me- 
ridian. The number of degrees on it between the 
star and the horizon will be the required meridian al- 
titude. 

1. What is the meridian altitude of Arcturus at De- 
troit? Ans. 67°. 

2. What is the meridian altitude of Vega in Lyra at 
Genoa? Ans. 82°. 

D2 



82 PROBLEMS ON THE GLOBES. 

3. What is the meridian altitude of Betelgeux in 
Orion at Bristol ? Ans. 44°. 

PROBLEM L. 

Given the latitude of a place, day of the month, and 
the altitude of a star, to find the hour of the night, 
and the star's azimuth. 

Rule. — Rectify the globe for the given latitude by 
Problem XV., and screw the quadrant of altitude to 
the celestial brazen meridian over the given latitude. 
Bring the sun's place in the ecliptic for the given day 
to the graduated side of the celestial brazen meridian, 
and set the hour circle to 12 o'clock. Turn the celes- 
tial globe on its axis westward until the star comes to 
the given altitude on the quadrant. The number of 
hours the index has passed over will show the time 
from noon when the star has the given altitude. The 
quadrant of altitude will intersect the horizon in the 
required azimuth. 

1. On the 2d of December, the star Deneb in Cygnus 
was observed at Paris to have an altitude of 27°. 
What was the hour of observation, and what the azi- 
muth of the star? Ans. 10 hours 30 minutes P.M.; 
azimuth, 52° west of north. 

2. At Dublin on July 20th, the star Altair in Aquila 
was observed to have an altitude of 43°. What was 
the hour of observation, and what the azimuth ? Ans. 
2 hours 15 minutes A.M. ; azimuth, 33° east of south. 

3. At Boston on May 8th, the star Antares in Scor- 
pio was observed to have an altitude of 10°. What 
was the time of observation, and what was the azimuth 
of the star? Ans. 10 hours 30 minutes P.M.; azi- 
muth, 59° east of south. 



PROBLEMS ON THE GLOBES. 83 



PROBLEM LI. 

The latitude of a place, the day of the month, and the 
hour of the day being given, to find the altitude of 
any star, and its azimuth. 

Rule. — Rectify the globe for the given latitude by 
Problem XV., and screw the quadrant of altitude to 
the celestial brazen meridian over the given latitude. 
Bring the sun's place in the ecliptic for the given day 
to the graduated side of the brazen meridian, and set 
the hour circle to 12 o'clock. If the given hour be be- 
fore noon, turn the celestial globe on its axis eastward, 
but if after noon, westward until the index has passed 
over as many hours as the given time wants of noon. 
Fasten the globe in this position by screwing the mill- 
ed screw at the northern extremity of the axis. Move 
the quadrant of altitude until the star coincides with 
the graduated side of the quadrant. The degrees on 
the quadrant from the horizon to the star will be the 
altitude, and the distance in degrees between the north 
or south points of the horizon and the graduated side 
of the quadrant will be the azimuth. 

1. What is the altitude and what the azimuth of 
Bellatrix in Orion on January 7th, at 10 P.M., in Ha- 
vre? Ans. Altitude, 46° ; azimuth, 0°. 

2. What is the altitude and what the azimuth of 
Regulus in Leo at 11 P.M. on January 12th at New 
Orleans? Ans. Altitude, 39° ; azimuth, 83° east of 
south. 

3. What is the altitude and what the azimuth of 
Dubhe in Ursa Major at 11 P.M. in Glasgow on April 
20th ? Ans. Altitude, 74° ; azimuth, 63° west of north. 



84 PROBLEMS OX THE GLOBES. 



PROBLEM LIL 



Tlie day of the month and the azimuth of a star at a 
certain place being given, to find the starts altitude 
and the hour of the night. 

Rule. — Rectify the globe for the latitude of the 
given place by Problem XV., and screw the quadrant 
of altitude to the celestial brazen meridian over that 
latitude. Find the sun's place in the ecliptic, and 
bring it to the graduated side of the celestial brazen 
meridian. Set the hour circle to 12 o'clock, and bring 
the lower end of the quadrant of altitude to coincide 
with the given azimuth on the horizon. Holding the 
quadrant in this position, turn the celestial globe on 
its axis until the given star comes to the graduated 
edge of the quadrant of altitude. The hours passed 
over by the index will be the required time from noon. 
The number of degrees on the quadrant from the hori- 
zon to the star will be the required altitude. 

1. At London, on September 10th, the azimuth of e 
Delphinse was 20 degrees from the south toward east : 
what was its altitude, and what the hour of the night ? 

2. On the 8th of October, at London, the azimuth of 
/3 Aurigse was 50 degrees from the north toward the 
west : what was the altitude, and what the hour of the 
night ? 

3. At New York, on the 17th of March, the azimuth 
of Regulus was 74° 1ST. of W. : what was the altitude, 
and what the hour of the night ? 



PROBLEMS ON THE GLOBES. 85 



PROBLEM LIIL 

A given star being on the meridian, and another given 
star at the eastern or western half of the horizon, to 
find the latitude of a place. 

Rule. — Bring the first given star to the meridian; 
keep the globe from revolving by screwing the milled 
head at the northern extremity of the axis ; elevate or 
depress the north pole until the second given star is 
brought to the eastern or western half of the horizon. 
The required latitude will correspond in degrees with 
the altitude of the pole. 

1 . When Altair was rising and Arcturus in the me- 
ridian, what was the latitude? Ans. 18° south lati- 
tude. 

2. What is the latitude of that place which will have 
Sirius in the meridian while Arcturus is rising ? Ans. 
49^° north latitude. 

3. What is the latitude of that place which will have 
Aldebaran on the meridian while Nekar in Bootes is 
rising? Ans. 47° north latitude. 

PROBLEM LIV. 

The latitude of a place, the day of the month, and two 
stars having the same altitude being given, to find the 
hour. 

Rule. — Rectify the globe for the given latitude by 
Problem XV., and screw the quadrant of altitude to 
the celestial brazen meridian over that latitude. Bring 
the sun's place in the ecliptic for the given day to the 
graduated side of the brazen meridian, and set the 
hour circle to 12 o'clock. Turn the Hobe on its axis 



86 PROBLEMS ON THE GLOBES. 

westward until the two given stars coincide with the 
altitude on the quadrant. The hours passed over by 
the index will be the time from noon to the required 
hour. 

1. At what hour at London will Menkar in the 
Whale's jaw and Aldebaran, on the 5th of January, 
have each 35° of altitude ? 

2. At what hour at Edinburg will Altair in the 
Eagle and £ in the tail of the Eagle have, on the 18th 
of November, each an altitude of 35° ? 

3. At what hour at Dublin, on May 15th, will Benet- 
nach in Ursa Major and y in Bootes have each 56° of 
altitude ? 

PROBLEM LV. 

Two stars having the same azimuth, and their respect- 
ive altitudes bei?ig given, to find the latitude. 

Rule. — Extend the graduated edge of the quadrant 
of altitude from one star to the other, so that each star 
may stand opposite its given altitude on the quadrant. 
Elevate or depress the north pole until that division 
of the quadrant marked is at the horizon : then will 
the required latitude correspond in degrees with the 
altitude of the pole. 

1. The altitude of Arcturus was observed to be 40°, 
that of Cor. Caroli 68°, and their common azimuth 71° 
from south toward east : what was the latitude ? Ans. 
51i° north. 

2. The common azimuth of Vega in Lyra and a in 
Hercules was 60°, and their respective altitudes 70° 
and 39-|° : what was the latitude ? 

3. The respective altitudes of a and y in Ursa Major 
were 40° and 29-^°, and their common azimuth 30°: re- 
quired the latitude. 



PROBLEMS ON THE GLOBES. 87 



PROBLEM LVI. 

The day of the month, and the hour when any known 
star rises or sets being given, to find the latitude of 
the place. 

Rule. — Bring the sun's place on the ecliptic for the 
given day to the graduated edge of the celestial brazen 
meridian, and set the hour circle to 1 2 o'clock. If the 
given time be before noon, turn the globe eastward, 
but if after noon, westward on its axis as many hours 
as the given time wants of noon. Elevate or depress 
the north pole of the globe until the given star is 
brought to the horizon. The required latitude will 
correspond in degrees with the altitude of the pole. 
If the south pole be elevated, the latitude will be south, 
but if the north pole, it will be north. 

1. In what latitude does Mirach in Bootes rise at 
half past twelve at night on December 10th? Ans. 
51J° north. 

2. In what latitude does Pollux rise at 9 P.M. on 
September 12th? 

3. In what latitude does Sirius rise at 2 A.M. on 
January 14th? 

PROBLEM LVII. 

To find that day of the year on which any given star 
passes the meridian at any given hour. 

Rule. — Bring the given star to the graduated edge 
of the meridian, and set the hour circle to twelve 
o'clock. If the given hour be before noon, turn the 
globe westward, but if after noon, eastward until the 
index has passed over as many hours as the given 



88 PROBLEMS ON THE GLOBES. 

hour wants of noon. The required day will be found 
on the ecliptic where it is crossed by the brazen me- 
ridian. 

1. On what day of the year does Procyon come to 
the meridian of London at 3 A.M. ? Ans. December 
1st. 

2. On what day of the month does Deneb in Leo 
come to the meridian of Washington at 9 P.M. ? 

3. On what day of the year does Bellatrix come to 
the meridian of New York at 11 P.M. ? 

PROBLEM LVIII. 

The day of the month being given, to find at what hour 
a given star will come to the meridian. 

Rule. — Bring the sun's place on the ecliptic for the 
given day to the graduated side of the brazen me- 
ridian, and set the hour circle to twelve o'clock. Turn 
the globe on its axis westward until the given star 
comes to the graduated side of the meridian. The 
hours passed over by the index will be the hour from 
noon when the given star comes to the meridian. 

1. At what hour does Regulus in Leo come to the 
meridian of London on September 23d ? Ans. 9h. 45m. 
A.M. 

2. At what hour will Menkar come to the meridian 
of New York on May 1st ? 

3. At what hour will Dubhe come to the meridian 
of New York on March 30th ? 



PROBLEMS ON THE GLOBES. b ( J 



PROBLEM LIX. 

The altitudes of two stars being given, to find the lati- 
tude of a place. 

Rule. — Subtract the latitude of each star from 90°, 
and with the remainders in degrees equal to those on 
the equinoctial, with the given stars as centres, de- 
scribe two arcs intersecting each other in the zenith. 
Bring this point of intersection to the graduated edge 
of the meridian, and that number on the meridian 
which coincides with the point of intersection will be 
the required latitude. 

Note. — To describe arcs on a glass globe, it is neces- 
sary to use a pencil made of hard soap. 

1. At sea, m north latitude, the altitude of Capella 
was observed to be 30°, and that of Aldebaran 35°. 
What was the latitude ? Ans. 20^° north. 

2. When the altitude of Markab in Pegasus was 30°, 
that of Altair in Aquila was 65 9 . What was the lati- 
tude in the northern hemisphere ? 

3. When the altitude of Procyon was observed to 
be 50°, that of Betelgeux in Orion was 58°. What was 
the latitude of the place of observation ? 

PROBLEM LX. 

Given the meridian altitude of any particular star, to 
find the latitude. 

Rule. — Find the declination of the given star, and 
bring a corresponding declination on the other side of 
the equinoctial to the horizon. The elevation of the 
pole will correspond with the required latitude in de- 
grees. If the north pole be elevated, the latitude will 



90 PROBLEMS ON THE GLOBES. 

be north, but if the south pole, the latitude will be 
south. 

1. In what degree of north latitude is the meridian 
altitude of Aldebaran 52 i°? Ans. 53 £°. 

2. In what degree of north latitude is the meridian 
altitude of Altair 90°? 

3. In what degree of north latitude is Regulus in the 
zenith when he culminates ? 

PROBLEM LXI. 

Given the latitude of a place, the day and hour, together 
icith the altitude and azimuth of a star, to find the 
star. 

Rule. — Rectify the globe for the latitude by Prob- 
lem XV., and screw the quadrant of altitude to the 
celestial brazen meridian over the given latitude. Find 
the sun's place in the ecliptic for the given day, and 
brinsr it to the Graduated side of the brazen meridian. 
Set the hour circle to twelve o'clock. If the given 
hour be before noon, turn the globe on its axis east- 
ward, but if after noon, westward until the index has 
passed over as many hours as equals the difference be- 
tween the given hour and noon. Secure the globe in 
this position by screwing the milled head at the north- 
ern extremity of the axle, and bring the division mark- 
ed on the quadrant to the given azimuth on the ho- 
rizon. Then under the given altitude on the quadrant 
will be found the required star. 

1. At London, on December 21st, at 4 o'clock in the 
morning, the altitude of a star was 50°, and its azimuth 
37° from the south toward the east. What was the 
name of the star ? Ans. Deneb in Leo. 

2. At London, on December 21st, at 4 o'clock A.M., 



PROBLEMS OX THE GLOBES. 91 

the altitude of a star was 8°, and its azimuth 51° from 
the south toward the west. What was the name of 
the star? 

3. At London, on September 1st, at 9 P.M., the alti- 
tude of a star was 47°, and its azimuth 73° from the 
south toward the east. What was the name of the 
star ? 



THE EXD. 



